Is sand biotic or abiotic

At a particular temperature, K = 4.1 ✕ 10−6 for the following reaction. 2 CO2(g) 2 CO(g) + O2(g) If 2.3 moles of CO2 is initiall

mestny [16]

Answer:

concentration of $[O_2]$ = 0.0124 = 12.4 ×10⁻³ M

concentration of $[CO]$ = 0.0248 = 2.48 ×10⁻² M

concentration of $[CO_2]$ = 0.4442 M

Explanation:

Equation for the reaction:

$2CO_2_{(g)$ ⇄ $2CO_{(g)$ + $O_2_{(g)$

Concentration of $CO_2_{(g)$ = $\frac{2.3}{4.9}$ = 0.469

For our ICE Table; we have:

$2CO_2_{(g)$ ⇄ $2CO_{(g)$ + $O_2_{(g)$

Initial 0.469 0 0

Change - 2x +2x +x

Equilibrium (0.469-2x) 2x x

K = $\frac{[CO]^2[O]}{[CO_2]^2}$

K = $\frac{[2x]^2[x]}{[0.469-2x]^2}$

$4.1*10^{-6}=\frac{2x^3}{(0.469-2x)^2}$

Since the value pf K is very small, only little small of reactant goes into product; so (0.469-2x)² = (0.469)²

$4.1*10^{-6} = \frac{2x^3}{(0.938)}$

$2x^3 =3.8458*10^{-6$

$x^3 =\frac{3.8458*10^{-6}}{2}$

$x^3=1.9229*10^{-6$

$x=\sqrt[3]{1.9929*10^{-6}}$

x = 0.0124

∴ at equilibrium; concentration of $[O_2]$ = 0.0124 = 12.4 ×10⁻³ M

concentration of $[CO]$ = 2x = 2 ( 0.0124)

= 0.0248

= 2.48 ×10⁻² M

concentration of $[CO_2]$ = 0.469-2x

= 0.469-2(0.0124)

= 0.469 - 0.0248

= 0.4442 M

3 0

2 years ago

Jodi reacts 6.087 grams of cobalt (mw=58.933) with a solution containing excess gold (iii) chloride and recovers 13.572 g of sol

Makovka662 [10]

The correct answer is reaction 2 most likely occurred in their experiment.

So the given two reactions are as follows:

1) Au³⁺(aq) + Co(s) --> Au(s) + Co³⁺(aq)

2) 2Au³⁺(aq) + 3Co(s) --> 2Au(s) + 3Co³⁺(aq)

Mass of Co = 6.087 g

Molar mass of Co = 58.933 g/mol

Moles of Co = $\frac{6.087 g}{58.933 g/mol}$ = 0.1032

Mass of Au = 13.572 g

Molar mass of Au = 196.967 g/mol

Moles of Au = $\frac{13.572 g}{196.967 g/mol}$ = 0.069

Now based on reaction 1) the molar ratio between Co and Au is 1:1

So as per this reaction 0.1032 moles of Co would produce 0.1032 mole of Au

Again based on reaction 2) the molar ratio between Co and Au is 3:2

So as per this reaction 0.1032 moles of Co would produce $\frac{2 x 0.1032}{3}$ mole of Au

or, As per the second reaction 0.1032 moles of Co would produce 0.0688 moles of Au

8 0

3 years ago

Does the attractive forces between the bonded atoms become weaker or stronger when a chemical bond breaks?

kicyunya [14]

The more electrons that are shared between two atoms, the stronger their bond will be. Covalent bonding is an example.

5 0

2 years ago

HELP HELP HELP!!!!

notka56 [123]

Metals and metalloids :)))

4 0

2 years ago

Read 2 more answers

It was found that compound contained 68.1% carbon, 13.7% hydrogen, and 18.2% oxygen by mass.

tekilochka [14]

A. The empirical formula is the simplest ratio of components making up a compound.

In 100 g of the compound

C H O

mass 68.1 g 13.7 g 18.2 g

number of moles

68.1 g / 12 g/mol 13.7 g / 1 g/mol 18.2 g / 16 g/mol

= 5.68 mol = 13.7 mol 1.14 mol

divide by the least number of moles - which is 1.14 mol

5.68 / 1.14 = 5.0 13.7 / 1.14 = 12.0 1.14 / 1.14 = 1.0

the ratio of the components C : H : O is 5: 12 : 1

therefore empirical formula is C₅H₁₂O

B. Molecular formula is the actual ratio of the components making up the compound. We need to find the mass of the empirical formula / empirical unit and then find how many empirical units make up the molecular formula

mass of 1 empirical unit - C₅H₁₂O - 12 g/mol x 5 + 1 g/mol x 12 + 16 g/mol x 1

= 60 + 12 + 16 = 88 g

number of empirical units = molecular formula mass / mass of 1 empirical unit

= 176.34 g/mol / 88 g = 2.00 = 2

there are 2 empirical formulas in the molecular formula

therefore molecular formula = 2 x (C₅H₁₂O) = C₁₀H₂₄O₂ ,

therefore molecular formula is C₁₀H₂₄O₂

3 0

2 years ago